Physics For Scientists And Engineers 6E - part 187

SECTION 24.2 • Gauss’s Law

745

Quick Quiz 24.3

If the net flux through a gaussian surface is zero, the follow-

ing four statements could be true. Which of the statements must be true? (a) There are no
charges inside the surface. (b) The net charge inside the surface is zero. (c) The
electric field is zero everywhere on the surface. (d) The number of electric field lines
entering the surface equals the number leaving the surface.

Quick Quiz 24.4

Consider the charge distribution shown in Figure 24.9. The

charges contributing to the total electric flux through surface S* are (a) q

only (b) q

only (c) q

and q

(d) all four charges (e) none of the charges.

Quick Quiz 24.5

Again consider the charge distribution shown in Figure

24.9. The charges contributing to the total electric field at a chosen point on the
surface S* are (a) q

only (b) q

only (c) q

and q

(d) all four charges (e) none of the

charges.

Conceptual Example 24.3 Flux Due to a Point Charge

A spherical gaussian surface surrounds a point charge q.
Describe what happens to the total flux through the
surface if

(A)

the charge is tripled,

(B)

the radius of the sphere is doubled,

(C)

the surface is changed to a cube, and

(D)

the charge is moved to another location inside the

surface.

Solution
(A) The flux through the surface is tripled because flux
is proportional to the amount of charge inside the surface.

(B) The flux does not change because all electric field lines
from the charge pass through the sphere, regardless of its
radius.

(C) The flux does not change when the shape of the
gaussian surface changes because all electric field lines
from the charge pass through the surface, regardless of its
shape.

(D) The flux does not change when the charge is moved to
another location inside that surface because Gauss’s law
refers to the total charge enclosed, regardless of where the
charge is located inside the surface.

another. The surface S* surrounds charges q

and q

; hence, the net flux through it is

)//

. Finally, the net flux through surface S 0 is zero because there is no charge

inside this surface. That is, all the electric field lines that enter S 0 at one point leave at
another. Notice that charge q

does not contribute to the net flux through any of the

surfaces because it is outside all of the surfaces.

Gauss’s law, which is a generalization of what we have just described, states that

the net flux through any closed surface is

(24.6)

where q

represents the net charge inside the surface and

E represents the electric

field at any point on the surface.

A formal proof of Gauss’s law is presented in Section 24.5. When using Equation

24.6, you should note that although the charge q

is the net charge inside the gaussian

surface,

E represents the total electric field, which includes contributions from charges

both inside and outside the surface.

In principle, Gauss’s law can be solved for

E to determine the electric field due to a

system of charges or a continuous distribution of charge. In practice, however, this type of
solution is applicable only in a limited number of highly symmetric situations. In the next
section we use Gauss’s law to evaluate the electric field for charge distributions that have
spherical, cylindrical, or planar symmetry. If one chooses the gaussian surface surround-
ing the charge distribution carefully, the integral in Equation 24.6 can be simplified.

E"d

A #

▲

PITFALL PREVENTION

24.1 Zero Flux is not Zero

Field

We  see  two  situations  in  which
there  is  zero  flux  through  a
closed  surface — either  there  are
no charged particles enclosed by
the  surface  or  there  are  charged
particles  enclosed,  but  the  net
charge inside the surface is zero.
For  either  situation,  it  is  incorrect
to conclude that the electric field
on the surface is zero. Gauss’s law
states that the electric flux is pro-
portional to the enclosed charge,
not the electric field.

Gauss’s law

746

CHAPTE R 24 • Gauss’s Law

Example 24.4 The Electric Field Due to a Point Charge

Starting with Gauss’s law, calculate the electric field due to
an isolated point charge q.

Solution A  single  charge  represents  the  simplest  possible
charge  distribution,  and  we  use  this  familiar  case  to  show
how  to  solve  for  the  electric  field  with  Gauss’s  law.  Figure
24.10 and our discussion of the electric field due to a point
charge  in  Chapter  23  help  us  to  conceptualize  the  physical
situation.  Because  the  space  around  the  single  charge  has
spherical  symmetry,  we  categorize  this  problem  as  one  in
which  there  is  enough  symmetry  to  apply  Gauss’s  law.  To
analyze any Gauss’s law problem, we consider the details of
the electric field and choose a gaussian surface that satisfies
some or all of the conditions that we have listed above. We
choose a spherical gaussian surface of radius r centered on
the point charge, as shown in Figure 24.10. The electric field
due to a positive point charge is directed radially outward by

symmetry and is therefore normal to the surface at every
point. Thus, as in condition (2),

E is parallel to dA at each

point. Therefore,

E " dA # E dA and Gauss’s law gives

By symmetry, E is constant everywhere on the surface, which
satisfies condition (1), so it can be removed from the inte-
gral. Therefore,

where we have used the fact that the surface area of a
sphere is 4(r

. Now, we solve for the electric field:

To finalize this problem, note that this is the familiar
electric field due to a point charge that we developed from
Coulomb’s law in Chapter 23.

What If?

What if the charge in Figure 24.10 were not at the

center of the spherical gaussian surface?

Answer In this case, while Gauss’s law would still be valid,
the situation would not possess enough symmetry to evalu-
ate the electric field. Because the charge is not at the center,
the magnitude of

E would vary over the surface of the

sphere and the vector

E would not be everywhere perpen-

dicular to the surface.

E #

4(/

dA # E

A # E

(4(

) #

E"d

A #

dA #

24.3 Application of Gauss’s Law to Various

Charge Distributions

As mentioned earlier, Gauss’s law is useful in determining electric fields when the
charge distribution is characterized by a high degree of symmetry. The following exam-
ples demonstrate ways of choosing the gaussian surface over which the surface integral
given by Equation 24.6 can be simplified and the electric field determined. In
choosing the surface, we should always take advantage of the symmetry of the charge
distribution so that we can remove E from the integral and solve for it. The goal in this
type of calculation is to determine a surface that satisfies one or more of the following
conditions:

1. The value of the electric field can be argued by symmetry to be constant over the

surface.

2. The dot product in Equation 24.6 can be expressed as a simple algebraic product

E dA because

E and d A are parallel.

3. The dot product in Equation 24.6 is zero because

E and d A are perpendicular.

4. The field can be argued to be zero over the surface.

All four of these conditions are used in examples throughout the remainder of this
chapter.

▲

PITFALL PREVENTION

24.2 Gaussian Surfaces

are not Real

A gaussian surface is an imaginary
surface that you choose to satisfy
the conditions listed here. It does
not have to coincide with a physi-
cal surface in the situation.

Figure 24.10 (Example 24.4) The point charge q is at the

center of the spherical gaussian surface, and

E is parallel to d A

at every point on the surface.

Gaussian

surface

SECTION 24.3 • Application of Gauss’s Law to Various Charge Distributions

747

Example 24.5 A Spherically Symmetric Charge Distribution

An insulating solid sphere of radius a has a uniform volume
charge density 1 and carries a total positive charge Q (Fig.
24.11).

(A)

Calculate the magnitude of the electric field at a point

outside the sphere.

Solution Because the charge distribution is spherically sym-
metric, we again select a spherical gaussian surface of radius
r, concentric with the sphere, as shown in Figure 24.11a. For
this choice, conditions (1) and (2) are satisfied, as they were
for the point charge in Example 24.4. Following the line of
reasoning given in Example 24.4, we find that

Note that this result is identical to the one we obtained for a
point charge. Therefore, we conclude that,

for a uniformly

charged sphere, the field in the region external to the

sphere is equivalent to that of a point charge located at

the center of the sphere.

(B)

Find the magnitude of the electric field at a point inside

the sphere.

Solution In this case we select a spherical gaussian surface
having  radius  r , a, concentric  with  the  insulating  sphere
(Fig.  24.11b).  Let  us  denote  the  volume  of  this  smaller
sphere  by  V *. To  apply  Gauss’s  law  in  this  situation,  it  is
important to recognize that the charge q

within the gauss-

ian surface of volume V * is less than Q. To calculate q

, we

use the fact that q

V *:

By symmetry, the magnitude of the electric field is
constant everywhere on the spherical gaussian surface and
is normal to the surface at each point — both conditions

* #

(

)

(for r - a)

(1)

E #

(1) and (2) are satisfied. Therefore, Gauss’s law in the
region r , a gives

Solving for E gives

Because

by definition and because k

1/4(/

this expression for E can be written as

Note that this result for E differs from the one we

obtained in part (A). It shows that E : 0 as r : 0.
Therefore, the result eliminates the problem that would
exist at r # 0 if E varied as 1/r

inside the sphere as it does

outside the sphere. That is, if E 2 1/r

for r , a, the field

would be infinite at r # 0, which is physically impossible.

What If?

Suppose we approach the radial position r $ a

from inside the sphere and from outside. Do we measure the
same value of the electric field from both directions?

Answer From Equation (1), we see that the field
approaches a value from the outside given by

From the inside, Equation (2) gives us

Thus, the value of the field is the same as we approach the
surface from both directions. A plot of E versus r is shown in
Figure 24.12. Note that the magnitude of the field is contin-
uous, but the derivative of the field magnitude is not.

E # lim

r : a

a # k

E # lim

r : a

(for r , a)

(2)

E #

4(/

1 #

Q /

(

E #

4(/

(

)

4(/

dA # E

dA # E (4(r

) #

(a)

Gaussian

sphere

(b)

Gaussian

sphere

Figure 24.11 (Example 24.5) A uniformly charged insulating

sphere of radius a and total charge Q. (a) For points outside the

sphere, a large spherical gaussian surface is drawn concentric

with the sphere. In diagrams such as this, the dotted line

represents the intersection of the gaussian surface with the

plane of the page. (b) For points inside the sphere, a spherical

gaussian surface smaller than the sphere is drawn.

E =

Figure 24.12 (Example 24.5) A plot of E versus r for a uniformly

charged insulating sphere. The electric field inside the sphere

(r , a) varies linearly with r. The field outside the sphere (r - a)

is the same as that of a point charge Q located at r # 0.

At the Interactive Worked Example link at http://www.pse6.com, you can investigate the electric field inside and outside
the sphere.

Interactive

748

CHAPTE R 24 • Gauss’s Law

Example 24.6 The Electric Field Due to a Thin Spherical Shell

A thin spherical shell of radius a has a total charge Q distrib-
uted uniformly over its surface (Fig. 24.13a). Find the
electric field at points

(A)

outside and

(B)

inside the shell.

Solution

(A) The calculation for the field outside the shell is identical
to that for the solid sphere shown in Example 24.5a. If we
construct a spherical gaussian surface of radius r - a concen-
tric with the shell (Fig. 24.13b), the charge inside this surface
is Q. Therefore, the field at a point outside the shell is equiv-
alent to that due to a point charge Q located at the center:

(B) The  electric  field  inside  the  spherical  shell  is  zero.  This
follows  from  Gauss’s  law  applied  to  a  spherical  surface  of
radius r , a concentric with the shell (Fig. 24.13c). Because of
the  spherical  symmetry  of  the  charge  distribution  and
because the net charge inside the surface is zero—satisfaction
of  conditions  (1)  and  (2)  again—application  of  Gauss’s
law shows  that  E # 0  in  the  region r , a.  We  obtain  the
same results  using  Equation  23.11  and  integrating  over  the
charge  distribution.  This  calculation  is  rather  complicated.
Gauss’s  law  allows  us  to  determine  these  results  in  a  much
simpler way.

(for r - a)

E #

Example 24.7 A Cylindrically Symmetric Charge Distribution

Find the electric field a distance r from a line of positive
charge of infinite length and constant charge per unit
length 3 (Fig. 24.14a).

Solution The symmetry of the charge distribution
requires that

E be perpendicular to the line charge and

directed outward, as shown in Figure 24.14a and b. To
reflect the symmetry of the charge distribution, we select a
cylindrical gaussian surface of radius r and length ! that is
coaxial with the line charge. For the curved part of this
surface,

E is constant in magnitude and perpendicular to

the  surface  at  each  point—satisfaction  of  conditions
(1) and  (2).  Furthermore,  the  flux  through  the  ends  of
the  gaussian  cylinder  is  zero  because

E is parallel to

these surfaces—the first application we have seen of
condition (3).

We take the surface integral in Gauss’s law over the

entire gaussian surface. Because of the zero value of

E " dA

for the ends of the cylinder, however, we can restrict our
attention to only the curved surface of the cylinder.

The total charge inside our gaussian surface is 3!.

Applying Gauss’s law and conditions (1) and (2), we find
that for the curved surface

The area of the curved surface is A # 2(r!; therefore,

(24.7)

Thus,  we  see  that  the  electric  field  due  to  a  cylindrically
symmetric  charge  distribution  varies  as  1/r,  whereas  the
field external to a spherically symmetric charge distribution
varies  as  1/r

. Equation 24.7 was also derived by integra-

tion of the field of a point charge. (See Problem 35 in
Chapter 23.)

E #

2(/

E (2(r

) #

E"d

A # E

dA # E

A #

Figure 24.13 (Example 24.6) (a) The electric field inside a uniformly charged

spherical shell is zero. The field outside is the same as that due to a point charge Q

located at the center of the shell. (b) Gaussian surface for r - a. (c) Gaussian surface

for r , a.

Gaussian

sphere

Gaussian

sphere

(a)

(c)

(b)

= 0

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